Answer
$Xe(g)+F_2(g)\longrightarrow XeF_2(g)$
$Xe(g)+2F_2(g)\longrightarrow XeF_4(s)$
$Xe(g)+3F_2(g)\longrightarrow XeF_6(s)$
Work Step by Step
Balance the equations individually such that there are the same number of atoms of each element in both the product and the reactants.